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What is the difference between an impulse and a step response diagram?

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If the impulse response is $h[n]$, $n \geq 0$, then the step response $s[m]$ at any time $m \geq 0$ is $$s[m] = h[0]+h[1]+h[2]+\cdots+h[m].$$ What other information do you want to get? The transfer function of the filter? That's easy: $$H(z) = h[0]+h[1]z^{-1} + h[2]z^{-2}+\cdots$$ –  Dilip Sarwate Apr 16 '13 at 20:26
    
So, because my filter was a second order filter then the step response is 2 times the impulse response! That makes also sense graphically. Thank you i understand it now. –  20317 Apr 16 '13 at 20:37
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Nope, the step response is the integral of the impulse response or the impulse response is the first derivative of the step response. That has nothing to do with filter order –  Hilmar Apr 16 '13 at 20:40
    
So in the graphic representation above the fact that the step response is a "double" version is due to integration of the impulse response? –  20317 Apr 16 '13 at 20:45
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It looks "double" because every second sample of the impulse response is zero. So, when integrating it, every second sample is exactly zero added to the previous value, thus it is the same. –  lxop Apr 16 '13 at 20:50

1 Answer 1

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If the impulse response is $h[n], n\geq 0$, then the step response $s[m]$ at any time $m≥0$ is $$s[m]=h[0]+h[1]+h[2]+\cdots +h[m].$$ Note that if the filter is a finite impulse response (FIR) filter so that $h[m] = 0$ for $m > M$, then the step response at $M$ is $$s[M] = h[0]+h[1]+h[2]+\cdots +h[M]$$ and remains at this value forever afterwards, that is, $s[m]=s[M]$ for all $m \geq M$.

What other information do you want to get? The transfer function of the filter? That's easy: $$H(z)=h[0]+h[1]z^{−1}+h[2]z^{−2}\cdots $$ which works out to be polynomial of degree $M$ in $z^{-1}$ for a FIR filter.

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